Question: A certain circle can be represented by the following equation. $x^2+y^2+4x-18y+81=0$ What is the center of this circle ? $($
Explanation: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2+4x-18y+81&=0\\\\ x^2+y^2+4x-18y&=-81\\\\ (x^2+4x)+(y^2-18y)&=-81 \text{(rearrange terms)}\\\\ (x^2+4x{+4})+(y^2-18y{+81})&=-81{+4}{+81}\end{aligned}$ Notice that we must add ${4}$ and ${81}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 4 and 81?] Writing the equation in standard form $\begin{aligned}(x^2+4x{+4})+(y^2-18y{+81})&=-81{+4}{+81}\\\\ (x+2)^2+(y-9)^2&=4\\\\ (x-(-2))^2+(y-9)^2&=2^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(-2,9)$ and has a radius of $2$ units. Summary The circle is centered at $(-2,9)$. The circle has a radius of $2$ units.